[seqfan] Digit "a" divides the concatenation "bc"

[Eric Angelini]

Hello SeqFans,

--let's call a, b and c any three consecutive digits
   of S (for instance 1, 2 and 3);
--S was build such that  "a" always divides the
   concatenation "bc" (here, 1 divides 23)

--two remarks:
(1) we accept the triplet (4,0,8), for instance,
    as we accept that < "4" divides "08" >;
(2) we never divide by zero (when "a" = 0 we
    simply ignore it).

--note:
Could the integer "100" (or 2000, or  700006,
etc., that is, integers having  2 or more consecutive
zeros) be part of S? Yes, as "1" (or "2" or "7") _do_
divide the concatenation "00".

--last point:
S was started with a(1)=1 and a(2)=2,  then
extended with the smallest integer not yet
present in S and not leading to a contradiction.

S = 1,2,3,4,5,6,54,52,50,51,53,57,56,540,8,
16,7,21,20,22,24,28,40,41,23,23,27,210,9,...

Why is "9" there ("9" is the last term of S computed
here)? Because after "210", the smallest available
integer not leading to a contradiction is 9:
--"2" divides "10"
--"1" divides any concatenation that start with "0x"
-- "0" is skipped
... We thus have a large choice for the integer
following "210" (and "9" will be itself followed by "18",
as "18" is the smallest available integer divisible by "9").

P.-S.
There is an infinite amount of integers that won't enter S.
I guess they form the seq T hereunder:

T=201,203,205,207,209,211,213,215,... ,301,302,
304,305,307,308,310,311,... ,401,402 403,405, etc.

Best,
É.